对0034.在排序数组中查找元素的第一个和最后一个位置中python代码块语法进行修正，并添加新解法。补充0203.移除链表元素使用原先链表的python解法

problems/0034.在排序数组中查找元素的第一个和最后一个位置.md

@@ -397,38 +397,74 @@ class Solution:
     def searchRange(self, nums: List[int], target: int) -> List[int]:
         def getRightBorder(nums:List[int], target:int) -> int:
             left, right = 0, len(nums)-1
-            rightBoder = -2 # 记录一下rightBorder没有被赋值的情况
+            rightBorder = -2 # 记录一下rightBorder没有被赋值的情况
             while left <= right:
                 middle = left + (right-left) // 2
                 if nums[middle] > target:
                     right = middle - 1
                 else: # 寻找右边界，nums[middle] == target的时候更新left
                     left = middle + 1
-                    rightBoder = left
+                    rightBorder = left
 
-            return rightBoder
+            return rightBorder
 
         def getLeftBorder(nums:List[int], target:int) -> int:
             left, right = 0, len(nums)-1 
-            leftBoder = -2 # 记录一下leftBorder没有被赋值的情况
+            leftBorder = -2 # 记录一下leftBorder没有被赋值的情况
             while left <= right:
                 middle = left + (right-left) // 2
                 if nums[middle] >= target: #  寻找左边界，nums[middle] == target的时候更新right
                     right = middle - 1
-                    leftBoder = right
+                    leftBorder = right
                 else:
                     left = middle + 1
-            return leftBoder
-        leftBoder = getLeftBorder(nums, target)
-        rightBoder = getRightBorder(nums, target)
+            return leftBorder
+        leftBorder = getLeftBorder(nums, target)
+        rightBorder = getRightBorder(nums, target)
         # 情况一
-        if leftBoder == -2 or rightBoder == -2: return [-1, -1]
+        if leftBorder == -2 or rightBorder == -2: return [-1, -1]
         # 情况三
-        if rightBoder -leftBoder >1: return [leftBoder + 1, rightBoder - 1]
+        if rightBorder -leftBorder >1: return [leftBorder + 1, rightBorder - 1]
         # 情况二
         return [-1, -1]
 ```
 ```python
+# 解法1的代码简化，直接使用left和right值作为左右边界标记
+class Solution:
+    def searchRange(self, nums: List[int], target: int) -> List[int]:
+        # 查找左边界（第一个等于 target 的位置）
+        def getLeftBorder(nums: List[int], target: int) -> int:
+            left, right = 0, len(nums) - 1
+            while left <= right:
+                middle = left + (right - left) // 2
+                if nums[middle] >= target:  # nums[middle] >= target 时，继续向左查找
+                    right = middle - 1
+                else:
+                    left = middle + 1
+            return left
+
+        # 查找右边界（最后一个等于 target 的位置）
+        def getRightBorder(nums: List[int], target: int) -> int:
+            left, right = 0, len(nums) - 1
+            while left <= right:
+                middle = left + (right - left) // 2
+                if nums[middle] > target:  # nums[middle] > target 时，继续向右查找
+                    right = middle - 1
+                else:
+                    left = middle + 1
+            return right
+
+        # 查找左边界和右边界
+        leftBorder = getLeftBorder(nums, target)
+        rightBorder = getRightBorder(nums, target)
+
+        # 判断目标值是否存在
+        if leftBorder <= rightBorder:
+            return [leftBorder, rightBorder]
+        else:
+            return [-1, -1]
+```
+```python
 # 解法2
 # 1、首先，在 nums 数组中二分查找 target；
 # 2、如果二分查找失败，则 binarySearch 返回 -1，表明 nums 中没有 target。此时，searchRange 直接返回 {-1, -1}；

problems/0203.移除链表元素.md

@@ -367,9 +367,32 @@ class Solution {
 ```
 
 ### Python：
+用原来的链表操作：
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
+        while head != None and head.val == val:
+            head = head.next
+        cur = head
+        while cur != None and  cur.next != None:
+            check = cur.next
+            if check.val == val:
+                cur.next = check.next
+                check = check.next
+            else:
+                cur = cur.next
+                check = check.next
+        return head
+
+```
 
 ```python
-（版本一）虚拟头节点法
+虚拟头节点法
 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, val=0, next=None):